Ex 7.3,1
ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that
ABD ACD
Given:
ABC is isosceles,
AB = AC
Also, DBC is isosceles,
DB = DC
To prove: ABD ACD
Proof :
In ABD and DBC, we have
AB = AC
BD = DC
AD = AD
ABD ACD
Ex 7.3,1
ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that
(ii) ABP ACP
From part (i),
ABD ACD
So, BAP = PAC
In ABP and ACP,
AB = AC
BAP = PAC
AP = AP
ABP ACP.
Ex 7.3,1
ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that
(iii)AP bisects as well as .
To prove: = & =
Proof:
From part (i) , ABD ACD
So, = .
Hence AP bisects A
For = ,
we will first prove BDP CDP
From part(ii) , ABP ACP
BP = CP
In BDP and CDP, we have
BD = CD
BP = CP
DP = DP
So, BDP CDP
=
Thus, AP bisects D
.
Ex 7.3,1
ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure ). If AD is extended to interest BC at P, show that
(iv) AP is the perpendicular bisector of BC
To prove: BP = CP & = = 90
Proof
From part(ii) , ABP ACP
BP = CP
=
Since BC is a line,
+ = 180
+ = 180
2 = 180
= (180 )/2
= 90
So, = = 90
Since, BP = CP
& = = 90
AP is perpendicular bisector of BC.

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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