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# Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry based on Measure of angle from AMC 10 A, 2014. You may use sequential hints to solve the problem.

## Measure of angle  - AMC-10A, 2019- Problem 13

Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$

,

• $90$
• $100$
• $105$
• $110$
• $120$

Geometry

Circle

Triangle

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-13

#### Check the answer here, but try the problem first

$110^{\circ}$

## Try with Hints

#### First Hint

According to the questation we draw the diagram. we have to find out $\angle BFC$

Now $\angle BEC$ = $\angle BDC$ =$90^{\circ}$ (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that $\angle ABC=70^{\circ}$ (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

#### Second Hint

As $\angle ABC=70^{\circ}$ and $\angle BEC=90^{\circ}$ Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is$180^{\circ}$

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

#### Third Hint

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$

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Try this beautiful Problem on Geometry based on Measure of angle from AMC 10 A, 2014. You may use sequential hints to solve the problem.

## Measure of angle  - AMC-10A, 2019- Problem 13

Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$

,

• $90$
• $100$
• $105$
• $110$
• $120$

Geometry

Circle

Triangle

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-13

#### Check the answer here, but try the problem first

$110^{\circ}$

## Try with Hints

#### First Hint

According to the questation we draw the diagram. we have to find out $\angle BFC$

Now $\angle BEC$ = $\angle BDC$ =$90^{\circ}$ (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that $\angle ABC=70^{\circ}$ (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

#### Second Hint

As $\angle ABC=70^{\circ}$ and $\angle BEC=90^{\circ}$ Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is$180^{\circ}$

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

#### Third Hint

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$

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