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Solution
Given BE = CD
When 2 sides of a triangle are equal then it is isosceles
When 2 angles and side of 2 angles are equal, then both triangles are simillar
∠ABP = ∠AQB (each 90°)
In ∆ ABE and ∆ ACD
BE = CD (Given)
∠BEA = ∠CDA (each 90°)
∠BAE = ∠CAD (Common angle)
=> ABE = ∠ACE (By angle sum property)
=> ∆ ABE is simillar to ∆ ACD
=> AB = AC
∴ The triangle is isosceles
Solution
Given: L is the bisector of an angle A and BT ⊥ AP and PQ
Proof: ∠BAP = ∠BAQ (L is the bisector of ∠A) -> given
∠ABP = ∠AQB (each 90°)
(given BP and BQ are perpendiculars from B to the area of angle A)
AB = BA (Common)
∴ ∆ APB ≅ ∆ AQB (∵ by ∆ AS)
=> BP = BQ (by CPCT)
or Point B is equidistant from area of ∆ A
Solution
√6 + √3 √6 - √3
√6 + √3 √6 - √3 x √6 + √3 √6 + √3
(√6 + √3)2 6 - 3
6 + 3 + 2√6 x √3 3
9 + 2√6x3 3
9 + 2√3x2x3 3
9 + 2 x 3√2 3
= 3 + 2√2
= a + b√2
=> a = 3 and b = 2
Solution
Given: Two parallel lines let l ∥ m
Let xy is a transversal intersecting l at p and m at Q.
PM and QN are bisectors of ∠XPL and QN is bisector of ∠PQM and ∠XPL = ∠PQM are corresponding ∠'s.
To prove: PM ∥ QN
Proof: ∠XPL = ∠PQM (Corresponding ∠'s as l ∥ m -> given)
=> ∠XPL / 2 = ∠PQM / 2
=> ∠1 = ∠3 (∵ PM is bisector of ∠XPL ∴ ∠1 = ∠2 and QN is bisector of ∠PQM ∴ ∠1 = ∠3)
∵ corresponding ∠'s are equal ∴ PM ∥ QN
Hence proved
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