Complete proofs involving triangles - High school Geometry Exercises

Find the missing step in the following reasoning in order to show that this triangle is isosceles.

Step
1	\widehat{A} +\widehat{B} +\widehat{C} =180^\circ
2	\widehat{B} =180^\circ-\left(75^\circ+30^\circ\right)=75^\circ
3	...
4	Conclusion : this triangle is isosceles

\widehat{B} =\widehat{C}

AB = AC

\widehat{B} +\widehat{C} =150^\circ

\widehat{A} and \widehat{B} are complementary angles

Find the missing step in the following reasoning in order to find the area of this triangle.

Step
1	S=\dfrac{1}{2}.BC.AH
2	Using the Pythagoras theorem we have: AH=\sqrt{{52}-36}=\sqrt{16}=4
3	...
4	S=\dfrac{1}{2}\times9\times4=18

Using the Pythagoras theorem we have: BH=\sqrt{{25}-16}=\sqrt{9}=3

The area of AHC equals 12

The area of AHB is equal to 6.

AH = BH

Find the missing step in the following reasoning in order to determine the measure of \widehat{B}

Step
1	Using the Pythagoras theorem we have: AH=\sqrt{25-9}=4
3	...
4	Conclusion : \widehat{B} = 30^\circ

\sin\left(\widehat{B}\right) =\dfrac{AH}{AB}= \dfrac{4}{8}= \dfrac{1}{2}

Using the Pythagoras theorem we have: BH=\sqrt{64-16}=\sqrt{48}

\widehat{B} +\widehat{C} =90^\circ

AB = BH

Find the missing step in the following reasoning in order to determine the length of \overline{BC}, given that the triangles are similar.

Step
1	We have \dfrac{DE}{AB}=2, the scale factor is 2.
2	We have \dfrac{DC}{BC}=\dfrac{CE}{AC}=2
3	...
4	Conclusion : BC=2

BC= 2DC

DC= 2BC

\widehat{B} =\widehat{D}

BC = \dfrac{1}{2}CE

Find the missing step in the following reasoning in order to determine the measure of \widehat{B}

Step
1	\widehat{a} + \widehat{BAC} +\widehat{b} =180^\circ
2	...
3	\widehat{B}+ \widehat{BAC} + \widehat{C}=180^\circ
4	\widehat{B}= 180^\circ - \left(50^\circ + 45^\circ\right) = 85^\circ

\widehat{BAC} =180^\circ-\left(70^\circ+65^\circ\right)=45^\circ

AB = AC

\widehat{B} +\widehat{C} =105^\circ

\widehat{BAC}=55^\circ

Find the missing step in the following reasoning in order to show that this triangle is equilateral.

Step
1	Since AB = AC then \widehat{A}= \widehat{B}
2	We have \widehat{B}= 60^\circ
3	...
4	Conclusion : this triangle is equilateral

\widehat{C} = 180^\circ-\left(60^\circ+60^\circ\right)= 60^\circ

AB = AC

\widehat{B} +\widehat{C} =120^\circ

\widehat{A}+\widehat{B}+\widehat{C}=180^\circ.

Find the missing step in the following reasoning in order to determine the area of ADE

Step
1	\widehat{A} +\widehat{B} +\widehat{C} =180^\circ
2	\widehat{A} =180^\circ-\left(90^\circ+60^\circ\right)=30^\circ
3	....
4	AE=2\sqrt{6}
5	Using the Pythagoras theorem we have AD=\sqrt{\left(2\sqrt{6}\right)^2-\sqrt{6}^2}=\sqrt{18}
6	S=\dfrac{1}{2} \times \sqrt{6} \times \sqrt{18}= \sqrt{27}

\dfrac{DE}{AE}=\sin\left(\widehat{A}\right) =\dfrac{1}{2}

\widehat{DEA}=60^\circ

BC=2\sqrt{6}

AE=2EC