Question

Given: AB ≅ BC , AE ≅ FC Prove: m∠AEC=m∠AFC

Answer 1

To prove m∠AEC = m∠AFC, we use the congruent segments AB ≅ BC and AE ≅ FC, and by the SSS congruence postulate, we show triangles AEB and CFB are congruent, leading to the conclusion by CPCTC that the angles m∠AEC and m∠AFC are equal.

Step-by-step explanation:

The student's question involves proving that two angles, m∠AEC and m∠AFC, are equal based on given conditions. This task requires an understanding of congruent segments and angle properties in geometry.

To prove that m∠AEC = m∠AFC:

1. Recognize that AB ≅ BC implies that line segment AB is congruent to line segment BC.
2. Similarly, AE ≅ FC means line segment AE is congruent to line segment FC.
3. Triangular congruence postulates such as Side-Angle-Side (SAS) could be used if there's an angle that is known to be congruent between the two triangles. But since that information is not given, we must rely on another property.
4. Since AE and FC are radii of the same circle, and AB and BC are congruent, triangles AEB and CFB are congruent by the SSS (Side-Side-Side) congruence postulate.
5. By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), it follows that ∠AEB is congruent to ∠CFB.
6. Therefore, m∠AEC which includes ∠AEB, and m∠AFC which includes ∠CFB, are equal by angle addition postulate (∠AEB and ∠CFB being congruent complements to the respective angles).

Answer 2

EC/AB = FC/MA

Step-by-step explanation:

We are given AB

≅

BC that means that side AB and side BC are equal also we know that angle opposite to equalt

sides are equal.

Hence, ∠BAE=∠BCE-------(1)

Also ∠AEB=∠CEB.

Now we are given that: ∠ABC = 130°30’ i.e. in degrees it could be given as:

60'=1°

30'=(1/2)°=0.5°

Hence ∠ABC = 130°30’=130+0.5=130.5°

Also we know that sum of all the angles in a triangle is equal to 180°.

Hence,

∠BAE+∠BCE+∠ABC=180°.

2∠BAE+130.5=180 (using equation (1))

2∠BAE=49.5

∠BAE=24.75° (DIVIDE BOTH SIDE BY 2)

Now in triangle ΔBEC we have:

∠BEC=90° , ∠BCE=24.75°

SO,

∠BEC+∠BCE+∠EBC=180°.

Hence, 90+24.75+∠EBC=180

∠EBC=180-(90+24.75)

∠EBC=65.25°

Now we are given AE = 10 in

Also ∠BEA= 90°.

And ∠BAE=24.75°; hence using trignometric identity to find the measure of side BE.

\begin{gathered}\tan 24.75=\dfrac{BE}{AE}=\dfrac{BE}{10}\\\\BE=10\tan 24.75-------(2)\end{gathered}tan24.75=AEBE=10BEBE=10tan24.75−−−−−−−(2)

similarly in right angled triangle ΔBEC we have:

\begin{gathered}\tan 24.75=\dfrac{BE}{EC}\\\\\tan 24.75=\dfrac{BE}{EC}\\\\EC=\dfrac{BE}{\tan 24.75}------(3)\end{gathered}tan24.75=ECBEtan24.75=ECBEEC=tan24.75BE−−−−−−(3)

Hence, using equation (2) in equation (3) we get:

EC=10 \text{in.}EC=10in.

Hence AC=AE+EC=10+10=20 in.

Hence side AC=20 in.