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liq [111]
3 years ago
13

ABC is an isosceles triangle in which ab and ac are equal if d is a the midpoint of BC prove that ABD=ADC​

Mathematics
1 answer:
jeka943 years ago
4 0

Answer:

Step-by-step explanation:

Given that,

ABC is an Isosceles triangle.

In an Isosceles triangle the opposite sides ( AB =AC) are equal; Their base angles ( < ABD = < ACD) are also equal to each other.

It is als given that D is the mid point of BC.

i.e., BD = CD

Therefore,

By SAS theorem of congruency of triangles,

ABD = ACD

If this is the answer required, hope it helps...

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Answer:

2 square cm

Step-by-step explanation:

Given :

A square is inscribed in a circle whose radius is r = 1 cm

Therefore, the diameter of the circle is 2 r = 2 x 1

                                                                      = 2 cm.

So the diagonal of the square is 2r.

Using the Pythagoras theorem, we find each of the side of the triangle is $r \sqrt 2$.

Therefore, the area of the square is given by $\text{(side)}^2$

                                                                         = $(r\sqrt 2)^2$

                                                                         $= 2 r^2$

                                                                         $= 2 (1)^2$

                                                                         $=2 \ cm^2$

Hence the area of the largest square that is contained by a circle of radius 1 cm is 2 cm square.

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