kadetskihykw

2022-05-02

In triangle ABC, $\mathrm{\angle}A=20\xba$ and $AM=CN=CB$ . Find angle $\mathrm{\angle}MBN$ ?.

Cristal Roth

Beginner2022-05-03Added 13 answers

Step 1

We draw an equilateral triangle with side AB as shown. Then $\mathrm{\u25b3}PAC$ is isosceles with $\mathrm{\angle}PAC={40}^{\circ}$ . That leads to $\mathrm{\angle}BPC={10}^{\circ}$ . Also, $\mathrm{\angle}PBC={20}^{\circ}$ .

Now $\mathrm{\u25b3}PBC\cong \mathrm{\u25b3}BAM$ (by S-A-S)

So it follows that $\mathrm{\angle}ABM={10}^{\circ}$ .

$\therefore \mathrm{\angle}MBN={80}^{\circ}-{50}^{\circ}-{10}^{\circ}={20}^{\circ}$

We draw an equilateral triangle with side AB as shown. Then $\mathrm{\u25b3}PAC$ is isosceles with $\mathrm{\angle}PAC={40}^{\circ}$ . That leads to $\mathrm{\angle}BPC={10}^{\circ}$ . Also, $\mathrm{\angle}PBC={20}^{\circ}$ .

Now $\mathrm{\u25b3}PBC\cong \mathrm{\u25b3}BAM$ (by S-A-S)

So it follows that $\mathrm{\angle}ABM={10}^{\circ}$ .

$\therefore \mathrm{\angle}MBN={80}^{\circ}-{50}^{\circ}-{10}^{\circ}={20}^{\circ}$

Friegordigh7r7

Beginner2022-05-04Added 16 answers

Step 1

Using angle chasing, $\mathrm{\angle}ACB=\mathrm{\angle}ABC=\frac{180\xba-20\xba}{2}=80\xba$ . Since $\mathrm{\Delta}NCB$ is isosceles, $\mathrm{\angle}BNC=\mathrm{\angle}NBC=50\xba$ .

Now we have to make use of the information $AM=CN=CB$ , focusing in on AM in particular. Thus, let us reflect triangle ABC horizontally around the middle, so that base AB stays in the same position. Thus $CN=CB=A{C}^{\prime}=AM$ . Since $\mathrm{\angle}{B}^{\prime}=80\xba$ by symmetry, $\mathrm{\angle}CAM=80\xba-20\xba=60\xba$ . Since $\mathrm{\Delta}CAM$ is also iscoceles, this triangle must be equilateral! Therefore $\mathrm{\angle}A{C}^{\prime}M=\mathrm{\angle}AM{C}^{\prime}=60\xba$ as well.

Again by symmetry, we have that $AE=EB$ . Thus $\mathrm{\angle}EBA=20\xba$ also and $\mathrm{\angle}AEB=140\xba$ . Now if we construct D such that $AM=MD$ and D lies on AB, then $\mathrm{\angle}MDA=20\xba$ as well. Thus by AA similarity, $\mathrm{\Delta}AMD\sim \mathrm{\Delta}AEB$ .

Using angle chasing, $\mathrm{\angle}ACB=\mathrm{\angle}ABC=\frac{180\xba-20\xba}{2}=80\xba$ . Since $\mathrm{\Delta}NCB$ is isosceles, $\mathrm{\angle}BNC=\mathrm{\angle}NBC=50\xba$ .

Now we have to make use of the information $AM=CN=CB$ , focusing in on AM in particular. Thus, let us reflect triangle ABC horizontally around the middle, so that base AB stays in the same position. Thus $CN=CB=A{C}^{\prime}=AM$ . Since $\mathrm{\angle}{B}^{\prime}=80\xba$ by symmetry, $\mathrm{\angle}CAM=80\xba-20\xba=60\xba$ . Since $\mathrm{\Delta}CAM$ is also iscoceles, this triangle must be equilateral! Therefore $\mathrm{\angle}A{C}^{\prime}M=\mathrm{\angle}AM{C}^{\prime}=60\xba$ as well.

Again by symmetry, we have that $AE=EB$ . Thus $\mathrm{\angle}EBA=20\xba$ also and $\mathrm{\angle}AEB=140\xba$ . Now if we construct D such that $AM=MD$ and D lies on AB, then $\mathrm{\angle}MDA=20\xba$ as well. Thus by AA similarity, $\mathrm{\Delta}AMD\sim \mathrm{\Delta}AEB$ .

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