Areas of Parallelograms and Triangles
Given: ABC is an isosceles triangle with AB = AC.
To Prove: ∠B = ∠C
Construction: Draw AP π BC
Proof: In right triangle APB and right triangle
APC,
Hyp. AB = Hyp. AC | Given
Side AP = Side AP | Common
∴ ∆APB ≅ ∆APC | RHS Rule
∴ ∠ABP = ∠ACP | C.P.C.T.
⇒ ∠B = ∠C.
Sponsor Area
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Line I is the bisector of an angle ∠A and B is any point on I. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
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