Answer:

Construct → Drawn DY parallel to BX

To prove →

EX

BE

=3

Proof :- In ΔADY using midpoint theorem

EX=

2

DY

also in ΔBCX using midpoint theorem

DY=

2

BX

∴EX=

4

BX

or, 4EX=BX

4EX=BE+EX

3EX=BE

EX

BE

=3

Answer:

solution

Given: ΔABC is isosceles triangle in which AB=AC, with a circle inscribed in a triangle.

To prove: BD=DC

Proof: AF and AE are tangents drawn to the circle from point A.

Since two tangents drawn to a circle from the same exterior point are equal, AF=AE=a

Similarly, BF=BD=b, and CD=CE=c

ΔABC is a isosceles triangle in which AB=AC

So, a+b=a+c

Thus, b=c

Therefore, BD=DC

Hence proved.

solution

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SIMILAR QUESTIONS

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In a △ABC, right angled at A. The radius of the inscribed circle is 2cm.Radius of the circle touching the side BC and also sides AB and AC produced is 15cm.The length of the side BC measured in cm is

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In Figure 5, a △ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of △ABC is 54 cm

2

, then find the lengths of sides AB and AC.

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A circle is inscribed in a ΔABC having sides BC,CA and AB,16cm,20cm and 24cm respectively as shown in the figure . Find AD,BE and CF.

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A triangle ABC is inscribed in a circle D,E, F are the middle point of the sides BC, CA and AB respectively. ∠CAD=θ.AD=x,BD=y.Then area of the ΔABC is equal to

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Let the in circle of ΔABC touches the sides BC, CA, AB, at A

1

,B

1

,C

1

respectively. The incircle of ΔA

1

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touches its side of B

1

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,C

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A

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and A

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at A

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respectively and so on.

Lim ∠A

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=

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