Question

In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD=BF.​

Answer 1

Answer:

AD=BF (Proved)

Step-by-step explanation:

Given that, ABC is a right-angled triangle with ∠ABC= 90°.

Also given, BCDE is a square on side BC of ΔABC and ACFG is a square on side AC of ΔABC.

We have to prove that AD=BF.

Now, in ΔBCF and ΔACD,

(i) BC= CD (As they are the sides of same square BCDF )

(ii) CF= AC (As they are the sides of same square ACFG )

Now, it is clear from the given diagram that ∠BCF= ∠ACF+∠BCA

                                                                                   =90° + ∠BCA ....(1)

( Since ACFG is a square, so, ∠ACF=90°)

Similarly, it is clear from the given diagram that ∠ACD= ∠DCB+∠BCA

                                                                                   =90° + ∠BCA ..... (2)

( Since BCDE is a square, so, ∠DCB=90°)

So, from equation (1) and (2),

(iii) ∠BCF= ∠ACD

Therefore, from condition (i), (ii), (iii), we conclude that

ΔBCF ≅ ΔACD.

Hence, BF=AD

⇒AD=BF

(Proved)

Answer 2

Answer:

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Step-by-step explanation:

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