Question

i want the answer of 7 and 8

Answer 1

solution 7
given. x=y. andAB=CB

in ABE and CBD

<x=<y ( given)
AB is equal to CB ( given)
And <B=<B (Common
hence triangle ABE is congruent to Triangle CBD

SO AE=CD (cpct)
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Answer 2

Here's your answer!!

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7) Solution:-

It's given that,
=> AB =BC
And < x = <y

Now,
In ∆ ABE ,we have :-
=> Exterior AEB = EBA +BAE
=> < y = EBA + BAE .........(1)

In ∆ BCD,we have
=> Exterior CDB = CBA +BCD
=> < x = CBA + BCD............(2)

Since,
<x = <y
So,
From equation (1) and (2) ,we get :-
=>EBA +BAE =CBA +BCD
=>BAE =BCD (Since EBA=CBA)

Therefore,
In ∆BCD and BAE ,we have :-
=> B = B (common)
=> BC =AB (Given )
=>BAE =BCD

By criteria (ASA),
=> ∆ BCD is congruent to BAE
Hence,
CF =AE ( c.p.c.t)
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2) Solution:-

In ∆APB and ∆AQB,
=> <APB =<AQB
=> < PAB =<QAB
=> AB =AB

By criteria (AAS),
∆APB is congruent to ∆AQB
And hence,
BP =BA (By c.p.c.t)
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Hope it helps you!! :)