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君1in figure, AFECΔGDB and 4L2. Prove that ΔADE ~AABC.2 |
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Answer» Solution: [FIGURE IS IN THE ATTACHMENT] Given: ∆FEC ∆FEC ≅ ∆GDB EC= BD ( by CPCT)..........(1) Given: ∠1= ∠2 AE AD ………….. …… ..(2)(Sides opposite to equal angles are equal)From eq 1 and 2AE/EC = AD/BADDE || BC(Converse of Basic proportionality Theorem)∠1 = ∠3 and ∠2 = ∠4[ Corresponding angles] In ∆ADE and ∆ABC∠A = ∠A (Common)∠1 = ∠3 (proved above)∠2 = ∠4 (proved above) ∆ADE ~ ∆ABC [ by AAA similarity criterion] Like if you find it useful! |
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