2 EPA 2 BAD = ARE see Fig. 7.223 Shorthal DAP ERP ADBE

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In Fig. 7.223, D is the mid-point of side BC and $A E ⊥ B C$ . If BC = a AC=b, AB=c, ED=z, AD = p and AE =h, prove that:

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(ii) $c^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(iii) $b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

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$ AB$ is a line segment and$ P$ is its mid-point. $ D$ and $ E$ are points on the same side of $ AB$ such that $ \angle BAD =\angle ABE$ and $ \angle EPA = \angle DPB$ (see Fig.) .

Show that (i) $ \Delta DAP\cong \Delta EBP$ (ii) $ AD = BE$

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See fig. In

△ A B C

, seg

A D ⊥

seg

B C

. Prove that:

{A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}

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AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that ∆DAP ≅ ∆EBP.

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(a)
In Fig.,

D

is a point on hypotenuse

A C

Δ A B C,

such that

B D ⊥ A C, D M ⊥ B C and D N ⊥ A B .

Prove that :

(i) D M^{2} = D N . M C

(b)
In Fig.,

D

is a point on hypotentenuse

A C

Δ A B C,

such that

B D ⊥ A C, D M ⊥ B C and D N ⊥ A B . Prove that :

(ii)

D N^{2} = D M \times A N

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