# In Figure, D is the Mid-point of Side Bc and Ae ⊥ Bc. If Bc = A, Ac = B, Ab = C, Ed = X, Ad = P and Ae = H, Prove that - Mathematics

In Figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED
= x, AD = p and AE = h, prove that:

(i) b^2 = p^2 + ax + a^2/4

(ii) c^2 = p^2 - ax + a^2/4

(iii) b^2 + c^2 = 2p^2 + a^2/2

#### Solution

We have, D as the mid-point of BC

(i) AC2 = AE2 + EC2

b2 = AE2 + (ED + DC)2           [By pythagoras theorem]

b2 = AD2 + DC2 + 2DC × ED

b^2=p^2+(a/2)^2+2(a/2)xxx      [BC = 2CD given]

rArrb^2=p^2+a^2/4+ax            .........(i)

(ii) In ΔAEB, by pythagoras theorem

AB2 = AE2 + BE2

⇒ c2 = AD2 − ED2 + (BD − ED)2          [By pythagoras theorem]

⇒ c2 = p2 − ED2 + BD2 + ED2 − 2BD × ED

rArrc^2=p^2+(a/2)^2-2(a/2)xx x

rArrc^2=p^2+a^2/4-ax                    .........(ii)

(iii) Add equations (i) and (ii)

b^2+c^2=p^2+a^2/4+ax + p^2+a^2/4-ax

b^2+c^2=2p^2+(2a^2)/4

b^2+c^2=2p^2+a^2/2

Is there an error in this question or solution?
Chapter 7: Triangles - Exercise 7.7 [Page 121]

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 7 Triangles
Exercise 7.7 | Q 24 | Page 121

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