Question 1: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC . If AD is extended to intersect BC at P, show that

- `ΔABD≅ ΔACD`
- `ΔABP≅ ΔACP`
- AP bisects ∠A as well as ∠D
- AP is perpendicular bisector of BC

**Answer:** In ΔABD and ΔACD

`AB=AC`

`BD=CD`

`AD=AD`

So, `ΔABD≅ ΔACD` (SSS rule)

In ΔABP and ΔACP

`AB=AC`

`AP=AP`

`∠ABP=∠ACP` (angles opposite to equal sides)

So, `ΔABP≅ ΔCP` (SAS rule)

Since `ΔABP≅ ΔACP`

So, `∠BAP=∠CAP`

So, AP is bisecting ∠BAC

Similarly, ΔBDP and ΔCDP can be proven to be congruent and as a result it can be proved that AP is bisecting ∠BDC

Question 2: AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects angle A.

**Answer:** This can be solved like previous question.

Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that:

- `ΔABM≅ ΔPQN`
- `ΔABC≅ ΔPQR`

**Answer:** In ΔABM and ΔPQN

`AB=PQ`

`AM=PN`

`BM=QN` (median bisects the base)

So, `ΔABM≅ ΔPQN`

In ΔABC and ΔPQR

`AB=PQ`

`BC=QR`

`AC=PR` (equal medians mean third side will be equal)

So, `ΔABC≅ ΔPQR`

Question 4: BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

**Answer:** In ΔAEB and ΔAFC

`BE=CE` (perpendicular)

`AB=BC` (hypotenuse)

So, `ΔAEB≅ ΔAFC`

Question 5: ABC is an isosceles triangle with AB = AC. Draw AD ┴ BC to show that ∠B = ∠C.

**Answer:** After drawing AD ┴ BC

In ΔADC and ΔADB

`AC=AB`

`AD=AD`

`∠ADC=∠ADB`

So, `ΔADC≅ ΔADB`

So, `∠ACD=∠ABC`

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