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2 years ago

Help pls will give brainliest

Mathematics

1 answer:

Tomtit [17]2 years ago

3 0

Answer:

1) AB

2) BC

3) AC

ABC

4) PA and OH

5) SO and CP

6) CA and SH

(I know I already answered this, but..........)

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Vaselesa [24]

Answer:

$minor\ arc\ BD=100^o$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

A circumscribed angle is the angle made by two intersecting tangent lines to a circle

In this problem

BC and CD are tangents to the circle

BC=CD ----> by the Two Tangent Theorem

That means

Triangle ABC and Triangle ADC are congruent

$m\angle BAC=m\angle DAC$

Find the measure of angle BAC

In the right triangle ABC

$m\angle BAC+m\angle BCA=90^o$

substitute given value

$m\angle BAC+40^o=90^o$

$m\angle BAC=90^o-40^o=50^o$

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$m\angle BAD=2m\angle BAC$

$m\angle BAD=2(50^o)=100^o$

Find the measure of minor arc BD

we know that

$minor\ arc\ BD=m\angle BAD$ -----> by central angle

therefore

$minor\ arc\ BD=100^o$

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