Question-198391

Question Number 198391 by mr W last updated on 19/Oct/23

Commented by mr W last updated on 19/Oct/23

prove that BE is bisector of ∠CBD. (B, E are touching points)

$${prove}\:{that}\:{BE}\:{is}\:{bisector}\:{of}\:\angle{CBD}. \\ $$$$\left({B},\:{E}\:{are}\:{touching}\:{points}\right) \\ $$

Answered by MM42 last updated on 19/Oct/23

A=(1/2)(CE^⌢ −BE^⌢ )=B_1 −C (i) B_1 +B_2 =(1/2)(BE^⌢ +BD^⌢ )=C+A (ii) (i),(ii)⇒B_1 =B_2 +B_3 ✓

$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({C}\overset{\frown} {{E}}−{B}\overset{\frown} {{E}}\right)={B}_{\mathrm{1}} −{C}\:\:\:\left({i}\right) \\ $$$${B}_{\mathrm{1}} +{B}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({B}\overset{\frown} {{E}}+{B}\overset{\frown} {{D}}\right)={C}+{A}\:\:\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow{B}_{\mathrm{1}} ={B}_{\mathrm{2}} +{B}_{\mathrm{3}} \:\:\checkmark \\ $$$$ \\ $$

Commented by MM42 last updated on 19/Oct/23