Mid-point and Its Converse (Including Intercept Theorem) — ICSE Class 9 Maths Important Questions

Answer: (b) $4.5$4.5 cm.

By the mid-point theorem $DE=\dfrac12 BC=\dfrac12\times9=4.5$DE=1/2 BC=1/2×9=4.5 cm.

Answer: (a) The mid-point of $PR$PR.

By the converse of the mid-point theorem, a line through the mid-point of one side parallel to another side bisects the third side, so $T$T is the mid-point of $PR$PR.

Answer: (c) Parallelogram.

Each mid-segment is parallel to and half a diagonal of the quadrilateral. Both pairs of opposite mid-segments become parallel and equal, so the figure is always a parallelogram.

Answer: (b) $3.2$3.2 cm.

By the intercept theorem, if parallel lines make equal intercepts on one transversal they make equal intercepts on any transversal. So the second intercept equals the first, $3.2$3.2 cm.

Answer: (a) By the mid-point theorem $EF=\tfrac12 BC$EF=1/2 BC, $FD=\tfrac12 CA$FD=1/2 CA, $DE=\tfrac12 AB$DE=1/2 AB. Adding, perimeter of $\triangle DEF=\tfrac12$△ DEF=1/2(perimeter of $\triangle ABC$△ ABC). R correctly explains A.

By the mid-point theorem, $DE\parallel BC$DE∥ BC and $DE=\dfrac12 BC$DE=1/2 BC.

Therefore $BC=2\times DE=2\times3.5=7$BC=2× DE=2×3.5=7 cm.

Relation: $DE$DE is parallel to $BC$BC and equal to half of $BC$BC.

Equal intercept theorem: if three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on every other transversal.

Since $AB=BC$AB=BC, $EF=DE=3$EF=DE=3 cm.

By the mid-point theorem:

$FD\parallel AC$FD∥ AC and $FD=\dfrac12 AC=AE=EC$FD=1/2 AC=AE=EC (since $E$E is mid-point of $AC$AC).

$DE\parallel AB$DE∥ AB and $DE=\dfrac12 AB=AF=FB$DE=1/2 AB=AF=FB (since $F$F is mid-point of $AB$AB).

$EF\parallel BC$EF∥ BC and $EF=\dfrac12 BC=BD=DC$EF=1/2 BC=BD=DC.

In $\triangle FBD$△ FBD and $\triangle DEF$△ DEF:
$FB=DE$FB=DE, $BD=EF$BD=EF, $FD=FD$FD=FD (common).

By SSS, $\triangle FBD\cong\triangle DEF$△ FBD≅ △ DEF.

In quadrilateral $AFDE$AFDE: $AF\parallel ED$AF∥ ED (as $DE\parallel AB$DE∥ AB) and $AE\parallel FD$AE∥ FD (as $FD\parallel AC$FD∥ AC). Both pairs of opposite sides are parallel, so $AFDE$AFDE is a parallelogram.

Construction: Join $AN$AN and produce it to meet $DC$DC produced at $P$P.

Proof: In $\triangle ABN$△ ABN and $\triangle PCN$△ PCN: $BN=CN$BN=CN ($N$N is the mid-point of $BC$BC), $\angle ANB=\angle PNC$∠ ANB=∠ PNC (vertically opposite), $\angle ABN=\angle PCN$∠ ABN=∠ PCN (alternate angles, $AB\parallel DP$AB∥ DP). By ASA, $\triangle ABN\cong\triangle PCN$△ ABN≅ △ PCN, so $AN=NP$AN=NP and $AB=CP$AB=CP.

In $\triangle ADP$△ ADP, $M$M is the mid-point of $AD$AD and $N$N is the mid-point of $AP$AP. By the mid-point theorem, $MN\parallel DP$MN∥ DP (so $MN\parallel DC\parallel AB$MN∥ DC∥ AB) and
$MN=\tfrac12DP=\tfrac12(DC+CP)=\tfrac12(DC+AB).$MN=1/2DP=1/2(DC+CP)=1/2(DC+AB). Hence proved.

Since $ABCD$ABCD is a parallelogram, $AB\parallel DC$AB∥ DC and $AB=DC$AB=DC, so $EB=\tfrac12AB=\tfrac12DC=DF$EB=1/2AB=1/2DC=DF and $EB\parallel DF$EB∥ DF. Hence $DEBF$DEBF is a parallelogram and $DE\parallel BF$DE∥ BF.

Let $DE$DE and $BF$BF meet $AC$AC at $P$P and $Q$Q.

In $\triangle ABQ$△ ABQ: $E$E is the mid-point of $AB$AB and $EP\parallel BQ$EP∥ BQ. By the converse of the mid-point theorem, $P$P is the mid-point of $AQ$AQ, so $AP=PQ$AP=PQ.

In $\triangle CDP$△ CDP: $F$F is the mid-point of $CD$CD and $FQ\parallel DP$FQ∥ DP. By the converse of the mid-point theorem, $Q$Q is the mid-point of $CP$CP, so $PQ=QC$PQ=QC.

Therefore $AP=PQ=QC$AP=PQ=QC: $DE$DE and $BF$BF trisect $AC$AC. Hence proved.

Given: $\triangle ABC$△ ABC in which $D$D and $E$E are the mid-points of $AB$AB and $AC$AC.

To prove: $DE\parallel BC$DE∥ BC and $DE=\dfrac12 BC$DE=1/2 BC.

Construction: Produce $DE$DE to $F$F so that $DE=EF$DE=EF, and join $CF$CF.

Proof: In $\triangle AED$△ AED and $\triangle CEF$△ CEF:

$AE=CE$AE=CE ($E$E is mid-point of $AC$AC)

$\angle AED=\angle CEF$∠ AED=∠ CEF (vertically opposite angles)

$DE=EF$DE=EF (by construction)

By SAS, $\triangle AED\cong\triangle CEF$△ AED≅ △ CEF.

Hence $AD=CF$AD=CF and $\angle ADE=\angle CFE$∠ ADE=∠ CFE (c.p.c.t.).

The equal alternate angles $\angle ADE=\angle CFE$∠ ADE=∠ CFE imply $AD\parallel CF$AD∥ CF, i.e. $DB\parallel CF$DB∥ CF.

Now $AD=CF$AD=CF and $AD=DB$AD=DB (as $D$D is mid-point of $AB$AB), so $DB=CF$DB=CF.

Thus $DB$DB and $CF$CF are equal and parallel, so $DBCF$DBCF is a parallelogram. Therefore $DF\parallel BC$DF∥ BC and $DF=BC$DF=BC.

Since $D,E,F$D,E,F are collinear, $DE\parallel BC$DE∥ BC.

Also $DF=DE+EF=2DE$DF=DE+EF=2DE, so $2DE=BC$2DE=BC, giving $DE=\dfrac12 BC$DE=1/2 BC. Hence proved.

Join diagonal $AC$AC.

In $\triangle ABC$△ ABC, $P$P and $Q$Q are mid-points of $AB$AB and $BC$BC. By the mid-point theorem:

$PQ\parallel AC \text{ and } PQ=\dfrac12 AC \quad\text{...(1)}$PQ∥ AC and PQ=1/2 AC ...(1)

In $\triangle ADC$△ ADC, $S$S and $R$R are mid-points of $AD$AD and $CD$CD. By the mid-point theorem:

$SR\parallel AC \text{ and } SR=\dfrac12 AC \quad\text{...(2)}$SR∥ AC and SR=1/2 AC ...(2)

From (1) and (2): $PQ\parallel SR$PQ∥ SR and $PQ=SR$PQ=SR.

Since one pair of opposite sides is equal and parallel, $PQRS$PQRS is a parallelogram.

Special case: Join diagonal $BD$BD. In $\triangle ABD$△ ABD, $PS=\dfrac12 BD$PS=1/2 BD; in $\triangle CBD$△ CBD, $QR=\dfrac12 BD$QR=1/2 BD. So $PS=QR=\dfrac12 BD$PS=QR=1/2 BD.

If $AC=BD$AC=BD, then $PQ=\dfrac12 AC=\dfrac12 BD=PS$PQ=1/2 AC=1/2 BD=PS. Thus two adjacent sides of the parallelogram are equal, so all four sides are equal.

Therefore, when the diagonals are equal, $PQRS$PQRS is a rhombus.

(i) By the mid-point theorem each mid-segment is half the opposite side:

$DE\parallel CA$DE∥ CA, $DE=\dfrac12 CA=\dfrac12\times100=50$DE=1/2 CA=1/2×100=50 m.

$EF\parallel AB$EF∥ AB, $EF=\dfrac12 AB=\dfrac12\times60=30$EF=1/2 AB=1/2×60=30 m.

$FD\parallel BC$FD∥ BC, $FD=\dfrac12 BC=\dfrac12\times80=40$FD=1/2 BC=1/2×80=40 m.

(ii) $EF$EF is parallel to $AB$AB and equal to half of $AB$AB.

(iii) Perimeter of $\triangle DEF=50+30+40=120$△ DEF=50+30+40=120 m.

(iv) Perimeter of $\triangle ABC=60+80+100=240$△ ABC=60+80+100=240 m. Fraction $=\dfrac{120}{240}=\dfrac12$=120/240=1/2, i.e. one half.