Page 6 - 10-Math-8 PROJECTION OF A SIDE OF A TRIANGLE

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Statements Reasons
In ∠rt rABD

(AB)2 = (AD)2 + (BD)2 (i) Pythagoras Theorem
In ∠rt rACD

(AC)2 = (AD)2 + (CD)2 (ii) Pythagoras Theorem

or (AC)2 = (AD)2 + (BD - BC)2 mBC + mCD = mBD

(AC)2 = (AD)2 + (BD)2 + (BC)2 - 2BC.BD (iii)

(AC)2 = (AB)2 + (BC)2 - 2BC.BD Using (i) and (iii)

Example 2: In an Isosceles rABC , if mAB = mAC and
then prove that (BC)2 = 2mAC .mCE
Given: In an Isosceles rABC

mAB = mAC and
whereas CE is the projection of BC upon AC .

To Prove: (BC)2 = mAC .mCE

Proof:

Statements Reasons

In an isosceles rABC with mAB = mAC if

∠C is acute.

Then (AB)2 = (AC)2 + (BC)2 - 2mAC .mCE By Theorem 2

(AC)2 = (AC)2 + (BC)2 - 2mAC .mCE Given mAB = mAC

⇒ (BC)2 - 2mAC .mCE = 0 Cancel (AC )2 on both
sides

or (BC)2 = 2mAC .mCE

EXERCISE 8.2

1. In a rABC, calculate mBC when mAB = 6cm, mAC = 4cm and m

∠A = 600

2. In a rABC 9cm and D is the mid

point of side AC . Find length of the median BD .

3. In a parallelogram ABCD, prove that

(AC)2 + (BD)2 = 2 [(AB)2 + (BC)2]

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