NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.3 – Part 1

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Corresponding Parts of Congruent Triangle(CPCT)

Give Corresponding Parts of Congruent Triangle If two triangles ABC and PQR are congruent under the corresponding A↔p,B↔Q and it is expressed as ΔABC ≅ ΔPQR

Give Corresponding Parts of Congruent Triangle

Give Corresponding Parts of Congruent Triangle If two triangles ABC and PQR are congruent under the corresponding A↔p,B↔Q and it is expressed as ΔABC ≅ ΔPQR

CPCT means that the corresponding sides are equal and the corresponding angles are equal.

Q-1 ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that

  1. ΔABDΔACD

  2. ΔABPΔACP

  3. AP bisects A as well as D.

  4. AP is the perpendicular bisector of BC.

Give ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC

Give ΔABC and ΔDBC Are Two Isosceles Triangles

Give ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC

Solution:

Given, ΔABC and ΔDBC are two isosceles triangles.

  1. In ΔABDΔACD AD=AD (Common line) AB=AC ( ΔABC is isosceles) BD=CD ( ΔDBC is isosceles)

    Therefore, ΔABDΔACD by SSS congruence condition

  2. In ΔABPΔACP AP=AP (Common line) PAB=PAC ( ΔABDΔACD so by CPCT) AB=AC ( ΔABC is isosceles)Therefore, ΔABPΔACP by SAS congruence condition.

  3. PAB=PAC by Corresponding Parts of Congruent Triangles as ΔABDΔACD .AP bisects A . ……………..equation (1)

    also,In ΔBPD and ΔCPD , PD=PD (Common line) BD=CD ( ΔDBC is isosceles.) BP=CP ( ΔABPΔACP so by Corresponding Parts of Congruent Triangle(CPCT)).

    Therefore, ΔBPDΔCPD by SSS congruence condition.

    Thus, BDP=CDP by CPCT. ………….equation(2)By (1) and (2) we can say that AP bisects A as well as D .

  4. BPD=CPD (by CPCT as ΔBPDΔCPD )and BP=CP ……………equation(3) also, BPD+CPD=180° (BC is a straight line.) 2BPD=180° BPD=90° ……………equation(3)

    From (1) and (2),

    AP is the perpendicular bisector of BC.

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