Q28 of 56 Page 6

In the given figure, points D and E trisect BC and ∠B = 90°. Prove that 8AE2 = 3AC2 + 5AD2.

Given: In Δ ABC, points D and E trisect on BC and ∠B = 90°.
To Prove: 8AE2 = 3AC2 + 5AD2.
Proof: Let ABC be the triangle in which h ∠ B = 90°. Let the points D and E trisect BC.
Join AD and AE. Then,
AC2 = AB2 + BC2
3AC2 = 3AB2 + 3BC2             ………..(i)
AD2 = AB2 + BD2
5AD2 = 5AB2 + 5BD2             ..……..(ii)
Therefore 3AC2 + 5AD2 = 8AB2 + 3BC2 + 5BD2
                                = 8AB2 + 3.+ 5
                                = 8AB2 +BE2
                                = 8AB2 + 8BE2
                                       = 8(AB2 + BE2)
                                = 8AE2.

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26 ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus. 27 Let ABC be a triangle, right-angled at C. If D is the mid-point of BC, prove that AB2 = 4AD2 – 3AC2.
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                                                      30  In fig,  ∠ B < 90° and segment AD ^ BC, show that b2 = h2 + a2 + x2 - 2ax