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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

18. angle CAB...

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18. angle CAB = 30 ,angle ACB = 90 . ACFG and BCDE are the squares of side AC and BC respectively. Segment BG and Segment AE intersect AC and BG at K

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Q. In triangle ABC, D and E are the points on sides AB and AC respectively. If AB = 18 cm, AD = 8 cm, AE = 12 cm and EC = 15 cm, then .

△ A B C, A B = A C

D

E

are on the sides

B C

A C

respectively such that

A D = A E

∠ B A D = 30^{\circ}

, then the measure of

∠ E D C

Q. In given figure,

E B ⊥ A C, B G ⊥ A E

C F ⊥ A E

Prove that:

(i)

Δ A B G \sim Δ D C B

\frac{B C}{B D} = \frac{B E}{B A}

If a line intersects sides AB and AC of a $△ A B C$ at D and E respectively and is parallel to BC, then $\frac{A D}{A B}$ = $\frac{A E}{x}$ Then $x$ is

Q. An isosceles triangle ABC has AC = BC. CD bisects AB at D and

∠

CAB = 55^o. Find

∠ A C B

.

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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

Standard IX Mathematics

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