Congurency In Triangles
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
So, `ΔABD≅ ΔACD` (SSS rule)
In ΔABP and ΔACP
`∠ABP=∠ACP` (angles opposite to equal sides)
So, `ΔABP≅ ΔCP` (SAS rule)
Since `ΔABP≅ ΔACP`
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
`BM=QN` (median bisects the base)
So, `ΔABM≅ ΔPQN`
In ΔABC and ΔPQR
`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
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
So, `ΔADC≅ ΔADB`