Question

Please solve this !!Q-2)

Answer 1

Hello!

• AB ⊥ BC and DM ⊥ BC
⇒ $\textbf{AB\: || \:DM}$

Similarly,

• CB ⊥ AB and DN ⊥ AB
⇒ $\textbf{ CB\: || \:DN}$

Hence, Quadrilateral BMDN is a rectangle.

∴ $\textbf{ BM = ND}$

♦ In ∆ $\textbf{BMD}$ :

∠1 + ∠BMD + ∠2 = 180°

⇒ ∠1 + 90° + ∠2 = 180°

⇒ ∠1 + ∠2 = 90°

Similarly,

♦ In ∆ $\textbf{DMC}$ :

∠3 + ∠4 = 90°

∠2 + ∠3 = 90°

Now,

∠1 + ∠2 = 90° = ∠2 + ∠3 = 90°

⇒ ∠1 + ∠2 = ∠2 + ∠3 ⇒ $\boxed{\angle \sf 1 = \angle \sf 3}$

Also,

∠3 + ∠4 = 90° = ∠2 + ∠3 = 90°

⇒ ∠3 + ∠4 = ∠2 + ∠3 ⇒ $\boxed{\angle \sf 2 = \angle \sf 4}$

Then,

In ∆ BMD & Δ DMC

• ∠1 = ∠3
• ∠2 = ∠4

Δ BMD ~ Δ DMC ---( AA Similarity Criterion )

$\dfrac{\sf BM}{\sf DM} = \dfrac{\sf MD}{\sf MC}$

⇒ $\dfrac{\sf DN}{\sf DM} = \dfrac{\sf MD}{\sf MC}$

⇒ $\boxed{\sf DM^{2} = DN\: x\:MC}$

Similarly,

Δ BND ~ Δ DNA ---( AA Similarity Criterion )

$\dfrac{\sf BN}{\sf DN} = \dfrac{\sf ND}{\sf NA}$

⇒ $\dfrac{\sf DM}{\sf DN} = \dfrac{\sf DN}{\sf AN}$

⇒ $\boxed{\sf DN^{2} = DM\: x\:AN}$

Cheers!

Answer 2

hope it helps you .....................

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