In the given Fig, ∠1 = ∠2. Prove that ∆ADE ~ ∆ABC.
Given: and
To Prove: Proof: In we have [given] AD = AE ...(i) [sides opposite equal angles are equal] [given] BD = CE ...(ii) [by CPCT] Dividing corresponding part of (i) by (ii), we get
Therefore, by converse of Basic proportionality theorem, DE || BC Now, in ∆ADE and ∆ABC ∠ADE = ∠ABC [corresponding angles] ∠AED = ∠ACB [corresponding angles] and ∠A = ∠A [common] Therefore, by using AAA condition ∆ADE ~ ∆ABC.
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304.
In the given Fig. DEFG is a square and ∠BAC = 90°. Prove that (i) ∆AGF ~ ∆DBG. (ii) ∆AGF ~ ∆EFC. (iii) ∆DBG ~ ∆AEFC (iv) DE2 = BD × EC.
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Short Answer Type
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307.
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309.
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