Question

please solve this question exercise 7.3 question number first class 9th Maths if you solve this question correctly I mark as brainlist answer

Answer 1

$\mathfrak{\blue{\huge{* hello *}}}$

$<b><i>$Given,

ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P...

$\mathcal{\large{\red{Proof :-}}}$

i} In ABD and ACD,

AB=AC [1] I ABC is an isosceles triangle...
BD=CD [2] I DBC is an isosceles triangle....
AD=AD [3] I common....

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ii} In ABP and ACP,

AB=AC [4] I from {i}...
L ABP=L ACP [5] I AD=AC from {1}.....
DABP=DACP [angles opposite to equal sides of a triangle are equal]......
ABP ~= ACP I proved in {i} above....
L BAP=L CAP {6} I CPCT....

In view of 4,5,6
ABP~= ACP.....

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iii} In BDP and CDP,

BD=CD [7] I from {2}...
DP=DP [8] I common...
ABP=ACP I proved in {ii} above...
\BP=CP [9] I CPCT...

In view of 7,8,9
BDP=CDP
L BDP=L CDP....
= DP bisect L D I SSS rule..
= AP bisect L D I CPCT....

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iv} BDP=CDP I proved in{iii} above...
\BP=CP [10] I CPCT....
L BPD=L CPD I CPCT..

But L BPD+ L CPD= 180° I linear pair axiom......
L BPD= L CPD= 90° [11]....

In view of 10 and 11...
AP is the perpendicular bisector of BC....

Answer 2

Answer:

Step-by-step explanation:

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