Question

1. triangleABC is a right triangle right angled at A such that AB = AC and
bisector of ZC intersects the side AB at D. Prove that AC + AD = BC.
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Answer 1

Step-by-step explanation:

Construct a triangle ABC where DE is perpendicular to BC Consider the △ DAC and △ DEC We know that ∠ BAC = ∠ DAC = 90o From the figure we know that CD bisects ∠ C So we get ∠ DCA = ∠ DCE We know that CD is common i.e. CD = CD By AAS congruence criterion △ DAC ≅ △ DEC So we know that DA = DE …… (1) AC = EC (c. p. c. t) ….. (2) It is given that AB = AC We know that the angles opposite to equal sides are equal ∠ B = ∠ C We know that the sum of angles of △ ABC is 180o. ∠ A + ∠ B + ∠ C = 180o By substituting the values 90o + ∠ B + ∠ B = 180o On further calculation 2 ∠ B = 180o – 90o 2 ∠ B = 90o By division ∠ B = 45o Considering the △ BED We know that ∠ BED = 90o So we can write it as ∠ BDE + ∠ B = 90o By substituting the values ∠ BDE + 45o = 90o On further calculation ∠ BDE = 90o – 45o By subtraction ∠ BDE = 45o It can be written as ∠ BDE = ∠ DBE = 45o We know that DE and BE are the equal sides of isosceles triangle DE = BE …….. (3) By comparing the equations (1) and (3) We get DA = DE = BE …… (4) We know that BC = BE + EC By considering the equations (ii) and (iv) We get BC = DA + AC We can also write it as AC + AD = BC Therefore, it is proved that AC + AD = BC.Read more on Sarthaks.com - https://www.sarthaks.com/718680/abc-right-triangle-right-angled-such-that-and-bisector-intersects-the-side-prove-that-ac-ad

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