Phần a,b nha 

a)Xét tứ giác MDHE, có:

MDHˆ=900MDH^=900

Mˆ=900M^=900

HEMˆ=900HEM^=900

=> Tứ giác MDHE là hình chữ nhật

b) Gọi giao điểm của MH là DE là O MDHE là hình chữ nhật nên hai đường chéo bằng nhau và cắt nhau tại trung điểm của mỗi đường

=> OH=OE

Xét tam giác EOH, có:

OH=OE(CMT)

=> Tam giác EOH cân tại O

=> H1ˆ=E1ˆH1^=E1^

Xét DEHP vuông tại E ,có:

A là trung điểm PH

=> AE = AH.

=> H2ˆ=E2ˆH2^=E2^

=> AEOˆ=AHOˆAEO^=AHO^ =900=900

Từ đó góc AEO = 900

hay tam giác DEA vuông tại E.

A)\(\text{T}ứ\&\text{nbsp};\text{gi} \overset{ˊ}{\text{a}} \text{c}\&\text{nbsp};\text{MDHE}\&\text{nbsp};\text{c} \overset{ˊ}{\text{o}} \&\text{nbsp};\text{ba}\&\text{nbsp};\text{g} \overset{ˊ}{\text{o}} \text{c}\&\text{nbsp};\text{vu} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{n} \hat{\text{e}} \text{n}\&\text{nbsp};\text{l} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{h} \overset{ˋ}{\imath} \text{nh}\&\text{nbsp};\text{ch}ữ\&\text{nbsp};\text{nh}ậ\text{t}.\)

B)\(\text{MDHE}\&\text{nbsp};\text{l} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{h} \overset{ˋ}{\imath} \text{nh}\&\text{nbsp};\text{ch}ữ\&\text{nbsp};\text{nh}ậ\text{t}\&\text{nbsp};\text{n} \hat{\text{e}} \text{n}\&\text{nbsp};\text{hai}\&\text{nbsp};đườ\text{ng}\&\text{nbsp};\text{ch} \overset{ˊ}{\text{e}} \text{o}\&\text{nbsp};\text{b} \overset{ˋ}{\overset{ }{\text{a}}} \text{ng}\&\text{nbsp};\text{nhau}\&\text{nbsp};\text{v} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{c} \overset{ˊ}{\overset{ }{\text{a}}} \text{t}\&\text{nbsp};\text{nhau}\&\text{nbsp};\text{t}ạ\text{i}\&\text{nbsp};\text{trung}\&\text{nbsp};đ\text{i}ể\text{m}\&\text{nbsp};\text{c}ủ\text{a}\&\text{nbsp};\text{m} \overset{\sim}{\hat{\text{o}}} \text{i}\&\text{nbsp};đườ\text{ng}.\)

\(\text{G}ọ\text{i}\&\text{nbsp};\text{O}\&\text{nbsp};\text{l} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{giao}\&\text{nbsp};đ\text{i}ể\text{m}\&\text{nbsp};\text{c}ủ\text{a}\&\text{nbsp};\text{MH}\&\text{nbsp};\text{v} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{DE}.\)

Ta có: OH = OE.=> góc H₁ = góc E₁

\(\text{DEHP}\&\text{nbsp};\text{vu} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{t}ạ\text{i}\&\text{nbsp};\text{E}\&\text{nbsp};\text{c} \overset{ˊ}{\text{o}} \&\text{nbsp};\text{A}\&\text{nbsp};\text{l} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{trung}\&\text{nbsp};đ\text{i}ể\text{m}\&\text{nbsp};\text{PH}\&\text{nbsp};\text{suy}\&\text{nbsp};\text{ra}:\&\text{nbsp};\text{AE}\&\text{nbsp};=\&\text{nbsp};\text{AH}.\)

=> góc H₂ = góc E₂

=> góc AEO và AHO bằng nhau mà góc AHO = 90⁰. 

\(\text{T}ừ\&\text{nbsp};đ \overset{ˊ}{\text{o}} \&\text{nbsp};\text{g} \overset{ˊ}{\text{o}} \text{c}\&\text{nbsp};\text{AEO}\&\text{nbsp};=\&\text{nbsp};\text{900}\&\text{nbsp};\text{hay}\&\text{nbsp};\text{tam}\&\text{nbsp};\text{gi} \overset{ˊ}{\text{a}} \text{c}\&\text{nbsp};\text{DEA}\&\text{nbsp};\text{vu} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{t}ạ\text{i}\&\text{nbsp};\text{E}.\)

C)DE = 2EA <=> OE = EA <=> tam giác OEA vuông cân  

<=> góc EOA = 45⁰ <=> góc HEO = 90⁰

<=> MDHE là hình vuông

<=> MH là phân giác của góc M mà MH là đường cao nên tam giác MNP vuông cân tại M.