Congurency In Triangles

Exercise 7.3

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

  1. `ΔABD≅ ΔACD`
  2. `ΔABP≅ ΔACP`
  3. AP bisects ∠A as well as ∠D
  4. AP is perpendicular bisector of BC
congurence triangles 2

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.

congurence triangles 4

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:

congurence triangles 6

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.

congurence triangles 8

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 angle B = angle C.

congurence triangles 10

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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